If m and n are integers such that -5

Question

Competition Mode Question

If m and n are integers such that -5 < m < 4 and -3 < n < 6. What is the maximum possible value of m^2 - mn + n^2 ?

A. 65
B. 61
C. 60
D. 59
E. 50

yashikaaggarwal · Answer

IMO Should Be B 
Let m = -4 
And n= 5
Putting values in the equation m^2-mn+n^2
We get,
(-4)^2-(-4)(5)+(5)^2
16+20+25
61

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NiftyNiffler · Answer

If m and n are integers such that -5 < m < 4 and -3 < n < 6. What is the maximum possible value of m^2 - mn + n^2 ?

A. 65
B. 61
C. 60
D. 59
E. 50

m = -4 , n = 5 will give us the maximum possible value = 16 + 20 + 25 = 61

Answer is  B

GMATWhizTeam · Answer

Solution 
 • m and n are integers
• -5 < m < 4
 o  -4 ≤ m ≤ 3  
•  -3 < n < 6 
 o  -2 ≤ n ≤ 5  
• We need to find the maximum value of  m^2 – mn + n^2 
 o Now,  m^2 – mn + n^2 = m^2 – mn + n^2 – mn + mn ⟹ (m – n)^2 + mn  
• So to find the maximum value of  (m – n)^2 + mn , we will check its value at extreme points of m and n.
 o We will try to maximize the value of  (m – n)^2 , as it has a power of 2, to maximize the value of the over all expression. 
• Extreme points of  (m, n)  are  (-4, -2), (-4, 5), (3, -2), (3, 5) 
 o We can observe that  (m -n)^2  will be maximum at  (-4, 5) 
o Thus, maximum value of  (m – n)^2 + mn = (-4 – 5)^2 – 4*5 = 81 – 20 = 61

Thus, the correct answer is  Option B.

exc4libur · Answer

(m-n)^2+mn=m^2 - mn + n^2
((-4)-(5))^2+(-4*5)
81-20
61

ans (B)

freedom128 · Answer

m and n are integers such that -4 <= m <= 3 and -2 <= n <= 5.

Maximum possible value is achieved when -m.n is maximum positive (m=-4, n=5)

m^2 - mn + n^2
= 16 + 20 + 25 = 61

FINAL ANSWER IS (B)

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MayankSingh · Answer

IMO B 
-5 < m < 4 and -3 < n < 6. 
m^2 - mn + n^2 = (m-n)^2 + mn

|m-n| max = |(-4)-(5)|= 9
So, (m-n)^2 + mn = 81- 20 = 61

B. 61

monikakumar · Answer

If m and n are integers such that -5 < m < 4 and -3 < n < 6. What is the maximum possible value of m^2 - mn + n^2 ?

A. 65
B. 61
C. 60
D. 59
E. 50

can be written as (m-n)^2+mn
so m=-4, n=5
81-20=61
Ans B

funkyakki · Answer

How do we know whether to subtract 1 from both of the equations -5 < m < 4 and -3 < n < 6 to arrive at the extreme points ?

ScottTargetTestPrep · Answer

Notice that m^2 - mn + n^2 = m^2 - 2mn + n^2 + mn = (m - n)^2 + mn. We have two cases:

1) Maximizing the value of (m - n)^2, i.e., having the values m and n as far as possible. However, this yields a negative value of mn since m and n must have opposite signs.

2) Maximizing the value of mn, i.e., having the values m and n as closes as possible. However, this yields a small nonnegative value of (m - n)^2.

If we use the first case, we should let m = -4 and n = 5 and (m - n)^2 + mn would be (-4 - 5)^2 + (-4)(5) = 81 - 20 = 61.

If we use the second option, we should let m = 3 and n = 6 and (m - n)^2 + mn would be (3 - 6)^2 + (3)(6) = 8 + 18 = 27.

Therefore, the maximum value is 61.

Answer: B

If m and n are integers such that -5 < m < 4 and -3 < n < 6. What is t

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GMAT Problem Solving (PS) Questions

If m and n are integers such that -5 < m < 4 and -3 < n < 6. What is t

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